Problem: This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.

This is an interactive problem.

Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.

The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by a⊕_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that x⊕_{k} z=y. Guess the password and break into the system.

Input

The first line of input contains a single integer t (1≤ t≤ 10 000) denoting the number of test cases. t test cases follow.

The first line of each test case contains two integers n (1≤ n≤ 2⋅ 10^5) and k (k=2).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Interaction

For each test case, first read two integers n and k. Then you may ask up to n queries.

For each query, print a single integer y (0≤ y≤ 2⋅ 10^7). Let the current password be x. After that, read an integer r.

If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.

Else, you will read r=0. At this moment the password is changed to a number z such that x⊕_{k} z=y.

After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.

To do this, use:

  * fflush(stdout) or cout.flush() in C++; 
  * System.out.flush() in Java; 
  * flush(output) in Pascal; 
  * stdout.flush() in Python; 
  * see documentation for other languages. 



If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.

Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.

Hacks:

To use hacks, use the following format of tests:

The first line should contain a single integer t (1≤ t≤ 10 000) — the number of test cases.

The first and only line of each test case should contain two integers n (1≤ n≤ 2⋅ 10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.

You must ensure that the sum of n over all test cases does not exceed 2⋅ 10^5.

Example

Input


1
5 2

0

0

1


Output


3

4

5

Note

In the example test case, the hidden password is 2.

The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2⊕_2 1=3.

The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1⊕_2 5=4.

The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.

Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class RPDAndRapSheet {
	public static void main(String[] args) throws IOException {
		int t = in.iscan();
		
		while (t-- > 0) {
			int n = in.iscan(), k = in.iscan();
			
			for (int i = 0; i < n; i ++) {
				out.println(i == 0 ? 0 : i ^ (i - 1));
				out.flush();
				int x = in.iscan();
				if (x == 1) break;
			}
		}
	}

	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
	}
}